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Hydrogen Spectral Series (Balmer Series)

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Matching Balmer series transitions with wavelengths

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2024 1 question

1 Atoms Medium

Match the List-I (spectral lines of Hydrogen for transitions from) with List-II (Wavelengths) below.

Pick the matching item from Column II for each row.
Column I
- A. $n_2=3$ to $n_1=2$ Match for A
- B. $n_2=4$ to $n_1=2$ Match for B
- C. $n_2=5$ to $n_1=2$ Match for C
- D. $n_2=6$ to $n_1=2$ Match for D
Column II
- I. 410.2 nm
- II. 434.1 nm
- III. 656.3 nm
- IV. 486.1 nm
Correct answer: A → III, B → IV, C → II, D → I

$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ Here $n_1 = 2$ is fixed (Balmer series), so as $n_2$ increases, $\frac{1}{\lambda}$ increases, i.e. the wavelength decreases. $n_2 = 3 \rightarrow 2$: 656.3 nm (longest) $n_2 = 4 \rightarrow 2$: 486.1 nm $n_2 = 5 \rightarrow 2$: 434.1 nm $n_2 = 6 \rightarrow 2$: 410.2 nm (shortest) Correct match: A-III, B-IV, C-II, D-I.